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To solve this problem, let's break it down step-by-step. 

### Problem Summary:
We are given a solid \( S \) where the cross-sections perpendicular to the x-axis are circles. The diameters of these circles extend from one curve to another:  
- From \( y = \frac{1}{2} \sqrt{x} \) (lower curve)  
- To \( y = \sqrt{x} \) (upper curve)  
For \( 0 \leq x \leq 4 \).

The radius of these circular cross-sections is given by the difference in the y-values of the two curves, meaning the diameter of the cross-section is the vertical distance between the curves at any given \( x \).

### Step 1: Expression for the Radius
The diameter of the circular cross-section at a point \( x \) is:
\[
\text{Diameter} = \sqrt{x} - \frac{1}{2} \sqrt{x} = \frac{1}{2} \sqrt{x}
\]

Thus, the radius \( r(x) \) of the circle at \( x \) is half the diameter:
\[
r(x) = \frac{1}{4} \sqrt{x}
\]

### Step 2: Area of Each Circular Cross-Section
The area of a circle with radius \( r(x) \) is:
\[
A(x) = \pi \left( r(x) \right)^2 = \pi \left( \frac{1}{4} \sqrt{x} \right)^2 = \pi \cdot \frac{1}{16} x = \frac{\pi}{16} x
\]

### Step 3: Volume of the Solid
To find the volume of the solid, we integrate the area of the cross-sections from \( x = 0 \) to \( x = 4 \):
\[
V = \int_0^4 A(x) \, dx = \int_0^4 \frac{\pi}{16} x \, dx
\]

### Step 4: Perform the Integration
The integral of \( x \) is:
\[
\int_0^4 x \, dx = \frac{x^2}{2} \Big|_0^4 = \frac{4^2}{2} = \frac{16}{2} = 8
\]

So, the volume becomes:
\[
V = \frac{\pi}{16} \times 8 = \frac{\pi}{2}
\]

### Final Answer:
The volume of the solid is \( \frac{\pi}{2} \) cubic units.

---

Would you like further details or clarifications on any step?

Here are 5 related questions to deepen understanding:
1. How would the solution change if the curves were different functions of \( x \)?
2. What is the impact of altering the bounds of integration from \( 0 \) to a different upper limit?
3. How does the volume change if we double the radius at each \( x \)?
4. Can you solve for the volume using the method of cylindrical shells instead of cross-sections?
5. How would you approach this problem if the cross-sections were ellipses instead of circles?

**Tip:** Always sketch the region or solid when dealing with cross-sections; it helps visualize the integration limits and geometry.

Cross-sections of the solid S in planes perpendicular to the x-axis are circles with diameters extending from the curve y = (1/2)√x to the curve y = √x for 0 ≤ x ≤ 4.

Explore the volume calculation of a solid formed by circular cross-sections with diameters extending from y = (1/2)√x to y = √x over the interval [0, 4]. The solution involves calculus concepts like definite integration and the area of a circle. Suitable for college-level students in Calculus I/II.

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